About a symbol $[ \ , L|K] : I_{K} \to G(L|K)$ (Algebraic Number Theory )

Question

I'am reading Jurgen Nueukirch's Algebraic number theory, p.387

I'm trying to understand the underlined statements (Here, $I_K$ is the idele group of $K$ (his book p.357) and ($\alpha _\mathfrak{p}, L_{\mathfrak{p}}|K_{\mathfrak{p}})$ is the local norm residue symbol (his book p.321))

Q.1) What does 'the completion $L_{p}$ with respect to $\mathfrak{P} | \mathfrak{p}$' means? What $\mathfrak{P}$ here means?

Q.2) Second, why almost all extensions $L_{p} | K_{p}$ are unramified? And how can we deduce that almost all factors in the product are $1$?

If $K$ is a number field and $v$ is a place, you can consider that $v$ (or $e^{-v}$ or some power of either) is a metric on $K$. The completion of $K$ wrt $v$ is simply the completion of $K$ as a metric space. It is a topological field with many good properties. In this case, $\mathfrak{P}$ is a place of $L$ extending the place $\mathfrak{p}$ of $K$ and you consider the completion of $L$ at said place. Its algebraic/topological structure doesn’t depend much on the choice of $\mathfrak{P}$ if $L/K$ is Galois because then the group acts transitively on the places of $L$ above $\mathfrak{p}$. — Aphelli, Jan 18 '22 at 09:16
(Cont’d) [place=equivalence class of nontrivial absolute values] for your second question, $L_p/K_p$ is unramified iff $L/K$ is unramified at $\mathfrak{P}$, which occurs at finitely many $\mathfrak{P}$ (use the discriminant, for instance). The last part is because the reciprocity symbol for local fields (or whatever Neukirch calls it) vanishes on the elements of the base fields that are norms, and because, for an unramified extension of local fields, every unit is a norm. — Aphelli, Jan 18 '22 at 09:21
Uhm. I trying to follow your comment. First, $\mathfrak{P}$ is arbitrary place extending $\mathfrak{p}$? And you said that "its algebraic/topological structure doesn’t depend much on the choice of $\mathfrak{P}$ if $L/K$ is Galois because then the group acts transitively on the places of $L$ above $\mathfrak{p}$" How this facts affects the formula for $[ , L|K]$ as product? Such a definition is really independent to $\mathfrak{P}$? — Plantation, Jan 18 '22 at 11:13
Second, you said that $L_{\mathfrak{p}}/K_{\mathfrak{p}}$ is unramified iff $L/K$ is unramified at $\mathfrak{P}$, which occurs at finitely many $\mathfrak{P}$. But correctly, "all but finitely many $\mathfrak{P}$", instead of " occurs at finitely many $\mathfrak{P}"? Any why this equivalence is true? Where can I find associated information? — Plantation, Jan 18 '22 at 11:19
Third, as you noted, the local norm residue symbol(Neukirch's book, p.321) $( \ , L|K) : K^{\ast} \to G(L|K)^{ab}$ has kernel $N_{L|K} L^{\ast}$. So key point is showing that every unit in the ring $\mathcal{O}$ of integers of $K$ is contained in $N_{L|K} L^{\ast}$. — Plantation, Jan 18 '22 at 11:26

score 1 · Accepted Answer · edited Aug 09 '24 at 05:55

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Although the comments address the question, I thought it might be useful to give an answer that explicitly links to the relevant parts of Neukirch's book. From corollary (2.12) in Chapter III, almost all prime ideals $\mathfrak{p}$ of $K$ are unramified in $L$, and for those $\mathfrak{p}$, $L_\mathfrak{P}|K_\mathfrak{p}$ is unramified for all places $\mathfrak{P}|\mathfrak{p}$. Then for these extensions $L_\mathfrak{P}|K_\mathfrak{p}$, if $\alpha_\mathfrak{p} \in K_\mathfrak{p}$ is a unit, then by proposition (6.2) in Chapter IV and theorem (1.1) in Chapter V it follows that $\alpha_\mathfrak{p} = N_{L_\mathfrak{P}|K_\mathfrak{p}}(y)$ for $y \in L_\mathfrak{P}$, in fact $y \in U_{L_\mathfrak{P}}$ ( Or use the (1.2) Corollary in Chapter V ). So from the definition of norm residue symbol on page 302, $(\alpha_\mathfrak{p}, L_\mathfrak{P}|K_\mathfrak{p}) = 1$ ( $\because$ $( \ ,L_\mathfrak{P}|K_\mathfrak{p} ) $ has kernel $N_{L_{\mathfrak{P}}|K_{\mathfrak{p}}}L_{\mathfrak{P}}^{*}$ ).

In case it's not clear that $\mathfrak{p}$ unramified in $L$ means $L_\mathfrak{P}|K_\mathfrak{p}$ unramified: the first indented equation on page 165 is $e_w = (w(L^*) : v(K^*))$. Since $\mathfrak{p}$ is unramified in $L$, $v_\mathfrak{p}(K^*) = v_\mathfrak{P}(L^*)$. From the third indented equation on page 126, $v_\mathfrak{p}(K_\mathfrak{p}^*) = v_\mathfrak{p}(K^*) = v_\mathfrak{P}(L^*) = v_\mathfrak{P}(L_{\mathfrak{P}}^*)$ so $L_{\mathfrak{P}}|K_\mathfrak{p}$ unramified. This is true no matter what $\mathfrak{P}|\mathfrak{p}$ is chosen.

edited Aug 09 '24 at 05:55

Plantation

3,710

answered Aug 01 '24 at 16:41

David Goldberg

417
3
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Thank you very much. Perhaps, can you explain why all choices of $\mathfrak{P} | \mathfrak{p}$ give the same result more in detail? For each $\alpha \in I_K$ with $\mathfrak{P}' , \mathfrak{P} | \mathfrak{p}$, $(\alpha_\mathfrak{p}, L_\mathfrak{P}|K_\mathfrak{p}) = (\alpha_\mathfrak{p}, L_\mathfrak{P'}|K_\mathfrak{p})$ ? And, why you mentioned the final paragraph in your answer? – Plantation Aug 02 '24 at 10:39
Perhaps there are two different questions (1) when $\mathfrak{p}$ is unramified, why is $(\alpha_\mathfrak{p}, L_{\mathfrak{P}i}|K\mathfrak{p}) = 1$ for every $\mathfrak{P}i|\mathfrak{p}$? (2) in general why is $(\alpha\mathfrak{p}, L_{\mathfrak{P}i}|K\mathfrak{p}) = (\alpha_\mathfrak{p}, L_{\mathfrak{P}j}|K\mathfrak{p})$? – David Goldberg Aug 04 '24 at 17:49
For (2), the idea is that when $L|K$ is abelian as it is here (or more generally Galois) then by Proposition (9.1) in Chapter I, there is a $\sigma \in G(L|K)$ that maps $\mathfrak{P}j$ to $\mathfrak{P}_k$ for any two $\mathfrak{P}_i|\mathfrak{p}$. Then $|\sigma x|{\mathfrak{P}k} = |x|{\mathfrak{P}j}$ (see proof of Prop (9.6) in Chapter II) so by continuity it extends to map $L{\mathfrak{P}j}$ isomorphically onto $L{\mathfrak{P}k}$ leaving $K\mathfrak{p}$ fixed, explaining why $L_{\mathfrak{P}j}$ and $L{\mathfrak{P}_k}$ have similar properties. – David Goldberg Aug 04 '24 at 17:49
For (1), here are the full details. Running things backwards: $(\alpha_\mathfrak{p}, L_\mathfrak{P}|K_\mathfrak{p}) = 1$ if $\alpha_\mathfrak{p} \in N_{L_\mathfrak{P}|K_\mathfrak{p}}L_{\mathfrak{P}}^*$. And that is true if $\alpha_\mathfrak{p}$ is a unit and $L_\mathfrak{P}|K_\mathfrak{p}$ is unramified. So it all boils down to understanding why $\mathfrak{p}$ unramified in $L$ means $L_{\mathfrak{P}i}|K\mathfrak{p}$ is unramified for every $\mathfrak{P}_i|\mathfrak{p}$. – David Goldberg Aug 04 '24 at 17:49
That follows from the very definition of unramified: $\mathfrak{p}$ unramified means that $\mathfrak{p} \mathcal{O}L = \mathfrak{P}_1 \cdots \mathfrak{P}_r$ where the $\mathfrak{P}_i$ consist of the entire set of $\mathfrak{P}|\mathfrak{p}$. That implies that when $v\mathfrak{p}(x) = 1$ then $v_{\mathfrak{P}i}(x) =1$ for every $i$. So $v{\mathfrak{P}i}| v{\mathfrak{p}}$ and since $v_{\mathfrak{P}i}(K^*) = v{\mathfrak{p}}(L^*) = \mathbb{Z}$ then as in the 2nd paragraph of my answer, that means $L_{\mathfrak{P}i}$ is unramified over $K\mathfrak{p}$. – David Goldberg Aug 04 '24 at 17:49
Thank you. Can I ask more? For (2), why $| \sigma x |{\mathfrak{P}_k} = | x|{\mathfrak{P}j}$ ? Can we prove that directly by using the definition of $| \cdot |{\mathfrak{P}k } $ and $ | \cdot |{\mathfrak{P}j}$? And why $\sigma$ extends to an isomorphism $L{\mathfrak{P}j} \to L{\mathfrak{P}k}$ leaving $K{\mathfrak{p}}$ fixed ? Is there any other reference we can refer to? – Plantation Aug 05 '24 at 06:11
For (1), What is the exact definition of unramifiedness of the extension of completions $L_{\mathfrak{P}{i}}/ K _{\mathfrak{p}}$ ( local fields )? In p.45 of the Neukirch's book, the extension $L{\mathfrak{P}{i}}/ K _{\mathfrak{p}}$ is called unramified if all prime ideals of $K{\mathfrak{p}}$ are unramifield in $L_{\mathfrak{P}{i}}$. But in which ring are we referring to the prime ideals? Its (discrete) valuation ring ? If we adopt this unramifiedness as definition, can we really show that $L{\mathfrak{P}_{i}}/ K _{\mathfrak{p}}$ is unramifield? – Plantation Aug 05 '24 at 06:11
I don't know why $(v_{\mathfrak{P}i}(L{\mathfrak{P}i}^{*}) : v{\mathfrak{p}}(K_{\mathfrak{p}}^{*})) =1 $ and $v_{\mathfrak{P}i} | v{\mathfrak{p}}$ leads to the unramifiedness of $L_{\mathfrak{P}{i}}/ K _{\mathfrak{p}}$. P.s. I also don't understand $v{\mathfrak{P}i} | v{\mathfrak{p}}$ rigorously. – Plantation Aug 05 '24 at 06:31
And..in the second paragraph of your anwer, you wrote that "it's not clear that $\mathfrak{p}$ unramified in $K$ means $L_{\mathfrak{P}}|K_{\mathfrak{p}}$ unramified." But, Perhaps, sentence "it's not clear that $\mathfrak{p}$ unramified in '$L$' means $L_{\mathfrak{P}}|K_{\mathfrak{p}}$ unramified.", is more correct? – Plantation Aug 05 '24 at 06:31
O.K. I think that I understood about the question $(2)$ ; https://math.stackexchange.com/questions/4954735/for-galois-extension-of-number-fields-mathfrakp-j-mathfrakp-k-mathfr . So it remains to understand the question $(1)$. Can you explain about it more friendly ? – Plantation Aug 06 '24 at 08:52
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Proposition: Suppose $L |K$ is a finite extension of number fields with rings $\mathcal{O}K$ and $\mathcal{O}_L$, $\mathfrak{p}$ is a prime of $\mathcal{O}_K$ and $\mathfrak{P}$ is a prime of $\mathcal{O}_L$ with $\mathfrak{P}|\mathfrak{p}$. Then if $\mathfrak{p}$ is unramified in $K$, the extension $L{\mathfrak{P}}|K_\mathfrak{p}$ is unramified. – David Goldberg Aug 09 '24 at 04:04
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Proof: To say that $\mathfrak{p}$ is unramified in $L$ means (page 49) $\mathfrak{p}\mathcal{O}L = \mathfrak{P}_1 \cdots \mathfrak{P}_r$. For any $x \in K$, write $x\mathcal{O}_K = \mathfrak{p}^k\mathfrak{q}$. The $\mathfrak{p}$-adic valuation $v\mathfrak{p}$ on $K$ is defined on page 165, giving $v_\mathfrak{p}(x) = k$. To compute $v_{\mathfrak{P}i}(x)$, $x\mathcal{O}_L = (x\mathcal{O}_K)\mathcal{O}_L = (\mathfrak{p}\mathcal{O}_L)^k \mathfrak{q}\mathcal{O}_L = \mathfrak{P}_1^k \cdots \mathfrak{P}_r^k \mathfrak{q}\mathcal{O}_L$, so $v{\mathfrak{P}_i}(x) = k$. – David Goldberg Aug 09 '24 at 04:05
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In other words $v_{\mathfrak{P}i} = v\mathfrak{p}$ on $K$ so $v_{\mathfrak{P}i}$ extends $v\mathfrak{p}$. Also, from the definition, $v_{\mathfrak{P}i}(L^*) = v\mathfrak{p}(K^) = \mathbb{Z}$. Applying the third indented equation on page 126, $v_{{\mathfrak{P}i}}(L{\mathfrak{P}_i}^) = v_{{\mathfrak{p}}}(K_{\mathfrak{p}}^*) = \mathbb{Z}$. But this means that $e(v_{\mathfrak{P}i}|v{\mathfrak{p}}) = 1$, (defined on page 150) which means that $L_{\mathfrak{P}i} | K{\mathfrak{p}}$ is unramified (definition (7.1) page 153 and Prop (8.5) on page 165). $\square$ – David Goldberg Aug 09 '24 at 04:05
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You might also want to refer to page 320, showing that the definition of ramification $e(L|K)$ in abstract valuation theory on page 285 matches the definition of ramification $e(w|v)$ for valuations on page 150. Note that for an extension of valued fields such as $L_{\mathfrak{P}i} | K{\mathfrak{p}}$, the ramification of that extension is defined to be the ramification of the valuations, $e(v_{\mathfrak{P}i} | v{\mathfrak{p}})$. – David Goldberg Aug 09 '24 at 04:06
Thank you. Q.1 ) For the first paragraph of your answer, why $(x\mathcal{O}_K)\mathcal{O}_L = (\mathfrak{p}^k \mathfrak{q})\mathcal{O}_L=(\mathfrak{p}\mathcal{O}_L)^k\mathfrak{q}\mathcal{O}_L$? And for the term $\mathfrak{P}_1^k \cdots \mathfrak{P}_r^{k}\mathfrak{q}\mathcal{O}_L$, why there is no $\mathfrak{P}_i$-term in the term $\mathfrak{q}\mathcal{O}_L$? – Plantation Aug 09 '24 at 05:05
Q. 2) And I still don't know how can we check the (7.1) Definition in p.153. The $L_{\mathfrak{P}i} | K{\mathfrak{p}}$ is finite extension of henselian fields? And why the extension $\lambda | \kappa$ of the residue class field is separable and $[L_{\mathfrak{P}i} : K{\mathfrak{p}} ] = [ \lambda : \kappa ]$? How can we use the Prop.(8.5) in page 165 ? – Plantation Aug 09 '24 at 05:06
O.K. For Q.1 ) I think it seems to be provable somehow. For Q. 2) $L_{\mathfrak{P}i}|K{\mathfrak{p}}$ is finite extension ( (8.4) Corollary in Chap. II ) of henselian fields ( p.131, (4.8) Theorem and p.147, (6.6) theorem ). So it remains to understand how can we check the definition (7.1) in p.153 for unramifiedness. – Plantation Aug 09 '24 at 09:16
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Q.2) Complete fields have only a single extension, so (8.5) on page 165 says $ef = [L:K]$. So definition (7.1) on page 153, which requires $f = [L:K]$ is equivalent to $e = 1$. – David Goldberg Aug 11 '24 at 06:09
O.K. Thank you. And, maybe finally, in checking the definition (7.1) in p.153, how can we show that the extension of residue fields $\lambda | \kappa$ is separable? – Plantation Aug 11 '24 at 09:15
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The residue class field of a local field is finite, and extensions of finite fields are always separable. – David Goldberg Aug 12 '24 at 17:22
Wow, O.K. I finally got it. I learned a lot. Thank you for your kindness ! – Plantation Aug 13 '24 at 05:36

About a symbol $[ \ , L|K] : I_{K} \to G(L|K)$ (Algebraic Number Theory )

1 Answers1